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Let $U$ be an open subset of $\mathbb{R}^m$ and $f:U\rightarrow \mathbb{R}^n$ a function. Fix $a\in U$.

Suppose $v$ is any non-zero vector in $\mathbb{R}^m$ and that $$L=\lim_{t\rightarrow 0} \frac{f(a+tv)-f(a)}{t}$$ exists and is independent of $v$ (i.e. it is same for all non-zero vectors $v$).

So directional derivative of $f$ at $a$ in all directions exists and is same.

Q. Can we conclude that $f$ is differentiable at $a$? If not, is it continuous at $a$?


The examples in books are given about existence of directional derivative in each direction but function is not continuous such as $f(x,y)=\frac{xy^2}{x^2+y^4}$ for $x\neq 0$ and $f(0,y)=0$. In this example, directional derivative of $f$ at $0$ exists along all directions; but its values are different. If $v=(a_1,a_2)$ then directional derivative is $a_2^2/a_1$ for $a_1\neq 0$, and its $0$ if $a_1=0$. So values of directional derivatives are different.

For the existence of derivative, we not only take linear directions to reach at $a$, but all curvilinear directions. I didn't find this question (mentioned above) in the books, stated as exercise at least, where above examples is mentioned.

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Define $f:\mathbb R^2\to \mathbb R$ by setting $f=1$ on the half parabola $\{(x,x^2): x>0 \},$ and $f=0$ everywhere else. Then all directional derivatives of $f$ at $(0,0)$ are $0.$ But $\lim_{x\to 0^+}f(x,x^2)=1\ne0= f(0,0).$ Thus $f$ is not even continuous at $(0,0),$ much less differentiable there.

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If $L=L(v)$ is independent of $v$, then $L\equiv 0$, since from the definition of $L$ we obtain that $L(-v)=-L(v)$.

Set $$ f(x,y)=\left\{\begin{array}{ccc} \dfrac{xe^{-1/y^2}}{x^2+e^{-2/y^2}} & \text{if} & y\ne 0 \\ 0 & \text{if} & y=0. \end{array} \right. $$ Then for all $b\ne 0$, $$ \frac{f(ta,tb)-f(0,0)}{t}=\dfrac{ae^{-1/(t^2b^2)}}{t^2a^2+e^{-2/(t^2b^2)}}= \dfrac{a}{t^2a^2e^{1/(t^2b^2)}+e^{-1/(t^2b^2)}}\to 0 $$ For $b=0$, the limit above is also zero. But $f$ is not even continuous at $(0,0)$.

Consider $$ \big(x(t),y(t)\big)=(ce^{-1/t^2},t)\to (0,0), \quad \text{as $t\to0$} $$ and $$ f\big(x(t),y(t)\big)=\frac{ce^{-2/t^2}}{(c^2+1)e^{-2/t^2}}=\frac{c}{c^2+1}\in [-1/2,1/2]. $$

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